Now if we are in a real vector space, which I suppose is assumed here, we have $p \cdot q = q \cdot p$. Note that in a complex vector space this does not hold, rather we have $p \cdot q = \overline{q \cdot p}$, the complex conjugate. Just mentioning this because commutativity of the dot product can not always be assumed.

However in this case we'll assume our vector space is over the real numbers so that we have $p \cdot r = q \cdot r \implies (p - q) \cdot r$ (by the bilinearity of the dot product) which shows that $p - q$ and $r$ are orthogonal.